Chemical formulas molar masses from

To determine the molar mass of a substance, we need its chemical formula and elemental molar masses. From the chemical formula, determine the number of moles of each element contained in one mole of the substance. Multiply each elemental molar mass by the number of moles of that element, and add. 

It is remarkable that no empirical mixture parameters and no experimental data are required to use the equation. The only parameters in the Flory-Huggins equation are the hard core volumes V, which are a pme-component property, and the atomic or group contribution values are found in standard compilations. Since the v/s are significant in the FH equation only in terms of their ratios, pure-liquid molar volumes are often used for V in place of hard core volumes. For solutions of polymers of the same chemical formula, molecular masses are legitimate substitutes for V , for the same reason. Thus the volume fractions ( ) can be substituted by mass fractions W . Either volume fraction or mass fraction is directly related to laboratory data. To avoid mole fractions, the activity tti from Equations (4.368) and (4.369) can be used to calculate by / = aj. ... 

V 132g (NH4)2S04 ) mo (NH4)2S04y molar mass from chemical formula... 

You will calculate molar masses from chemical formulas (Section 7.4). 

Sometimes chemists have to analyze substances about which they know very little. A chemist may isolate an interesting molecule from a natural source, such as a plant or an insect. Under these conditions the chemical formula must be deduced from mass percentage data, without the help of an expected formula. A four-step procedure accomplishes this by using mass-mole conversions, the molar masses of the elements, and the fact that a chemical formula must contain integral numbers of atoms of each element. 

To get from elemental analysis to a chemical formula, we begin by dividing each mass by the appropriate molar... 

Each element has a specific symbol that is different from the symbol for any other element. In a chemical formula, the symbol stands for an atom of an element. Molecular substances are composed of two or more atoms that are tightly bound together. The formula for a molecular substance consists of the symbols for the atoms that are found in that molecule. For instance, the formula for carbon dioxide is CO2. Note the use of the subscript to show that each molecule contains two oxygen atoms in addition to the one carbon atom. Also note that the 1 for the one carbon atom is not written. The molecular mass of CO2 is the sum of the atomic mass of carbon plus twice the atomic mass of oxygen and is expressed in u. As was discussed directly above, the molar mass of CO2 is the mass in grams equal to the molecular mass in u. A mole of carbon dioxide is 12.0 u + 2(16.0 u) = 44 u. This result can be expressed as 44 g to indicate one Avogadro s number, Na, of CO2 molecules. Recall that Na is 6.0221 x 1023 things—molecules in this case. 

If you know the chemical formula of any compound, then you can calculate the percentage composition. From the subscripts, you can determine the mass contributed by each element and add these to get the molar mass. Then, divide the mass of each element by the molar mass. Multiply by 100 to find the percentage composition of that element. 

You are given 35.6 g AICI3 and must calculate the number of Al + ions, the number of Cl ions, and the mass in grams of one formula unit of AICI3. Molar mass, Avogadro s number, and ratios from the chemical formula are the necessary conversion factors. The ratio of AP+ ions to CE ions in the chemical formula Is 1 3. Therefore, the calculated numbers of ions should be in that ratio. The mass of one formula unit in grams should be an extremely small number. 

Percent composition from the chemicai formuia If you already know the chemical formula for a compound such as water (H2O), can you calculate its percent composition The answer is yes. You can use the chemical formula to calculate the molar mass of water (18.02 g/mol) and assume you have an 18.02-g sample. Because the percent composition of a compound is always the same, no matter the size of the sample, you can assume that the sample... 

Concentration in solution usually initial (total) concentration or concentration range is reported and the same units (molar or mass concentration of certain compound or element) are used as in the original publications, The concentration can be easily converted from one unit to another when the chemical formula of the adsorbate is known. In a few original publications the molal concentration of the adsorbate was reported (mol per kg of solution). These concentrations were converted into molar concentration assuming that the specific density of the solution was 1 kg/dm . 

Each substance is identified by the usual form of its chemical formula and by its most commonly employed chemical name. In some cases, with in-termetallic compounds in particular, the name corresponds to a list of the chemical elements in the formula. The molar masses (grnol ) are calculated from the relative atomic masses of the elements taking A = 2 for C and are reported at the beginning or end of the first line. 

The molar mass of a compound can be calculated from its chemical formula and can be used to convert from mass to moles of that compound. 

Conversions between mass, moles, and the number of particles are summarized in Figure 10.11. Note that molar mass and the inverse of molar mass are conversion factors between mass and number of moles. Avogadros number and its inverse are the conversion factors between moles and the number of representative particles. To convert between moles and the number of moles of atoms or ions contained in the compound, use the ratio of moles of atoms or ions to 1 mole of compound or its inverse, which are shown on the upward and downward arrows in Figure 10.11. These ratios are derived from the subscripts in the chemical formula. 

Analyze We are asked to calculate the density of a gas given its name, its pressure, and its temperature. From the name we can write the chemical formula of the substance and determine its molar mass. 

Write down chemical formulae for the addition polymers formed from the following monomers, and in each case calculate the relative molar mass of a potymer having a degree of polymerization of 1000 ... 

When a new molecule is synthesized, an elemental analysis is routinely performed to help verify its identity. This test, which measures the mass percentage of each element in the compound, is also Ifequendy done as part of the process of identifying any substance whose composition is unknown. The mass percentages describe the compound s composition, and so they must be related to its chemical formula. But the data obtained from elemental analysis describe the composition in terms of the mass of each element, whereas the formula describes the composition in terms of the number of atoms of each element. So these are two different representations of very similar information, and the molar masses of the elements provide a connection between them. The process of obtaining the empirical formula of a compound from its percent composition by mass is best illustrated by an example. 

Chemical formulas are expressed in terms of numbers of particles. In this problem, we have to start with masses and somehow infer information about the numbers of particles. We did a similar problem in this chapter the determination of an empirical formula from percentage by mass data. In this case, with the experiment described, we can determine the mass information for each element in the unknown compound—providing the same type of data we had from percentage mass calculations. Then, because we know the empirical formula and we know the molar mass of the oxygen, we can determine the molar mass of the unknown metal. 

Knowing the chemical formula and the molecular mass of a compound enables us to calculate the percent composition by mass—the percent by mass of each element in a compound. It is useful to know the percent composition by mass if, for example, we needed to verify the purity of a compound for use in a laboratory experiment. From the formula we could calculate what percent of the total mass of the compound is contributed by each element. Then, by comparing the result to the percent composition obtained experimentally for our sample, we could determine the purity of the sample. Mathematically, the percent composition is obtained by dividing the mass of each element in 1 mole of the compound by the molar mass of the compound and multiplying by 100 percent ... 

The proportions of the various elements, n. n. n. n. n, must be rounded to the nearest integer to obtain the following empirical chemical formula C Hj,N O,S. Therefore, the relative molecular mass of the fuel, denoted M, can be easily calculated from the relative atomic molar mass of each chemical element, A, using the following equation ... 

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